What is the pH of (i) 1.0 X 103 M HBr and (ii) 1.0 X 102 M KOH?

Fundamentals Of Analytical Chemistry
9th Edition
ISBN:9781285640686
Author:Skoog
Publisher:Skoog
Chapter21: Potentiometry
Section: Chapter Questions
Problem 21.9QAP
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What is the pH of (i) 1.0 X 103 M HBr and (ii) 1.0 X 102 M KOH?

Expert Solution
Step 1

We know that, pH and pOH can be expressed as the negative logarithm of hydrogen ion and hydroxide ion respectively, i.e.

           pH = -log[H+]

 and, pOH = -log[OH-]

Also, pH + pOH = 14

Part (a)

We have, [HBr] = 1.0 x 10-3 M

Since, HBr is an strong electrolyte, hence it dissociate completely to give H+  and Br-.

Thus, [H+] = [HBr] = 1.0 x 10-3 M

So, pH = -log[H+]

            = -log(1.0 x 10-3)

           =  3

Therefore, the pH of the solution is 3.

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